A uniform rod of length `lambda` lies on a smooth horizontal table A particle moving on the table has a mass m and a speed v before the collision and it sticks to the rod after the collision. The rod has a mass M then find out.
The moment of inertia of the system about the vertical axis passing through the centre of mass C after the collision.

Explanation

Figure shows the situation of system just before and just after collision.
Initially the centre of mass of the rod is at point O. After collision when the particle sticks to the rod. Centre of mass is shifted from point O to C as shown in figure. Now the system is rotated about axis passing through C
Now from linear momentum conservation
` mv = (M + m)v' rArr v' = (mv)/(M + m) `
Let us assume that moment of inertia of the system
about c is . 1 . Then ` l = l_(("rod") C) +l_(("part")C)`
` l = l_(0) + Ml_(2)^(2) + ml_(1)^(2)`
` l = (Ml^(2))/(12) + (M m^(2) l^(2))/(4 (m + M)^(2)) + (m M^(2) l^(2))/(4(m + M)^(2))`
`rArr l = (M(M + 4m))/(12(m + M)) l^(2)`